Question

23. Show that (tan A x sin A) + cos A = sec A​

Answer 1

Given to prove :-

(tanA × sinA ) + cosA = secA

To know :-

${tanA} = \dfrac{sinA}{cosA}$

${secA} = \dfrac{1}{cosA}$

Solution :-

(tanA × sinA ) + cosA = secA

Take LHS

(tanA × sinA ) + cosA

${tanA} = \dfrac{sinA}{cosA}$

$\dfrac{sinA}{cosA} \times sinA + cosA$

$\dfrac{sin^2A }{cosA} + cosA$

Take LCM cosA

$\dfrac{sin^2A +cos^2A}{cosA}$

From , Trigonmetric Identities

sin²A + cos²A = 1

$\dfrac{1}{cosA}$

secA

hence LHS = RHS

PROVED !

know more :-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj